Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 11}{x - 6} = \dfrac{-x + 31}{x - 6}$
Multiply both sides by $x - 6$ $ \dfrac{x^2 - 11}{x - 6} (x - 6) = \dfrac{-x + 31}{x - 6} (x - 6)$ $ x^2 - 11 = -x + 31$ Subtract $-x + 31$ from both sides: $ x^2 - 11 - (-x + 31) = -x + 31 - (-x + 31)$ $ x^2 - 11 + x - 31 = 0$ $ x^2 - 42 + x = 0$ Factor the expression: $ (x - 6)(x + 7) = 0$ Therefore $x = 6$ or $x = -7$ At $x = 6$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 6$, it is an extraneous solution.